Question

In ‘Thinking in C++’ by Bruce Eckel, there is a program given to print a double value
in binary. (Chapter 3, page no. 189)

int main(int argc, char* argv[]) 
{  
  if(argc != 2) 
  {  
    cout << "Must provide a number" << endl;  
    exit(1);  
  }  
  double d = atof(argv[1]);  
  unsigned char* cp = reinterpret_cast<unsigned char*>(&d);  
  for(int i = sizeof(double); i > 0 ; i -= 2)   
  {  
    printBinary(cp[i-1]);  
    printBinary(cp[i]);  
  }
}

Here while printing cp[i] when i=8 (assuming double is of 8 bytes), wouldn’t it be undefined behaviour?
I mean this code doesn’t work as it doesn’t print cp[0].

Answer 1

A1: Yes, it would be undefined behaviour when it accesses cp[8].

A2: Yes, it also does not print cp[0].

As shown, it prints bytes 7, 8, 5, 6, 3, 4, 2, 1 of the valid values 0..7. So, if you have copied the code correctly from the book, there is a bug in the book’s code. Check the errata page for the book, if there is one.

It is also odd that it unwinds the loop; a simpler formulation is:

for (int i = sizeof(double); i-- > 0; )
    printBinary(cp[i]);

There is also, presumably, a good reason for printing the bytes in reverse order; it is not obvious what that would be.